49th International Mathematical Olympiad Spain

if and only if the diagonals $AC$ and $BD$ are perpendicular.

For a point $P$ in $ABCD$ which satisfies (1), let $K,L,M,N$ be the feet of perpendiculars from $P$ to lines $AB,BC,CD,DA$, respectively. Note that $K,L,M,N$ are interior to the sides as all angles in (1) are acute. The cyclic quadrilaterals $AKPN$ and $DNPM$ give $$\angle PAB+\angle PDC=\angle PNK+\angle PNM=\angle KNM.$$ Analogously, $\angle PBC+\angle PAD=\angle LKN$ and $\angle PCD+\angle PBA=\angle MLK$. Hence the equalities (1) imply $\angle KNM=\angle LKN=\angle MLK=90^{\circ }$, so that $KLMN$ is a rectangle. The converse also holds true, provided that $K,L,M,N$ are interior to sides $AB,BC,CD,DA$.

i. Suppose that there exists a point $P$ in $ABCD$ such that $KLMN$ is a rectangle. We show that $AC$ and $BD$ are parallel to the respective sides of $KLMN$. Let $O_A$ and $O_C$ be the circumcentres of the cyclic quadrilaterals $AKPN$ and $CMPL$. Line $O_A{O}_C$ is the common perpendicular bisector of $LM$ and $KN$, therefore $O_A{O}_C$ is parallel to $KL$ and $MN$. On the other hand, $O_A{O}_C$ is the midline in the triangle $ACP$ that is parallel to $AC$. Therefore the diagonal $AC$ is parallel to the sides $KL$ and $MN$ of the rectangle. Likewise, $BD$ is parallel to $KN$ and $LM$. Hence $AC$ and $BD$ are perpendicular.



ii. Suppose that $AC$ and $BD$ are perpendicular and meet at $R$. If $ABCD$ is a rhombus, $P$ can be chosen to be its centre. So assume that $ABCD$ is not a rhombus, and let $BR<DR$ without loss of generality. Denote by $U_A$ and $U_C$ the circumcentres of the triangles $ABD$ and $CDB$, respectively. Let $A{V}_A$ and $C{V}_C$ be the diameters through $A$ and $C$ of the two circumcircles. Since $AR$ is an altitude in triangle $ADB$, lines $AC$ and $A{V}_A$ are isogonal conjugates, i.e. $\angle DA{V}_A=\angle BAC$. Now $BR<DR$ implies that ray $A{U}_A$ lies in $\angle DAC$. Similarly, ray $C{U}_C$ lies in $\angle DCA$. Both diameters $A{V}_A$ and $C{V}_C$ intersect $BD$ as the angles at $B$ and $D$ of both triangles are acute. Also $U_A{U}_C$ is parallel to $AC$ as it is the perpendicular bisector of $BD$. Hence $V_A{V}_C$ is parallel to $AC$, too. We infer that $A{V}_A$ and $C{V}_C$ intersect at a point $P$ inside triangle $ACD$, hence inside $ABCD$. Construct points $K,L,M,N,O_A$ and $O_C$ in the same way as in the introduction. It follows from the previous paragraph that $K,L,M,N$ are interior to the respective sides. Now $O_A{O}_C$ is a midline in triangle $ACP$ again. Therefore lines $AC,O_A{O}_C$ and $U_A{U}_C$ are parallel. The cyclic quadrilateral $AKPN$ yields $\angle NKP=\angle NAP$. Since $\angle NAP=\angle DA{U}_A=\angle BAC$, as specified above, we obtain $\angle NKP=\angle BAC$. Because $PK$ is perpendicular to $AB$, it follows that $NK$ is perpendicular to $AC$, hence parallel to $BD$. Likewise, $LM$ is parallel to $BD$. Consider the two homotheties with centres $A$ and $C$ which transform triangles $ABD$ and $CDB$ into triangles $AKN$ and $CML$, respectively. The images of points $U_A$ and $U_C$ are $O_A$ and $O_C$, respectively. Since $U_A{U}_C$ and $O_A{O}_C$ are parallel to $AC$, the two ratios of homothety are the same, equal to $\lambda =AN/AD=AK/AB=A{O}_A/A{U}_A=C{O}_C/C{U}_C=CM/CD=CL/CB$. It is now straightforward that $DN/DA=DM/DC=BK/BA=BL/BC=1-\lambda$. Hence $KL$ and $MN$ are parallel to $AC$, implying that $KLMN$ is a rectangle and completing the proof.

Solution 2:

For a point $P$ distinct from $A,B,C,D$, let circles ($APD$) and ($BPC$) intersect again at $Q$ ($Q=P$ if the circles are tangent). Next, let circles ($AQB$) and ($CQD$) intersect again at $R$. We show that if $P$ lies in $ABCD$ and satisfies (1) then $AC$ and $BD$ intersect at $R$ and are perpendicular; the converse is also true. It is convenient to use directed angles. Let $\measuredangle (UV,XY)$ denote the angle of counterclockwise rotation that makes line $UV$ parallel to line $XY$. Recall that four noncollinear points $U,V,X,Y$ are concyclic if and only if $\measuredangle (UX,VX)=\measuredangle (UY,VY)$. The definitions of points $P,Q$ and $R$ imply $$\begin{array}{rl}\measuredangle (AR,BR)& =\measuredangle (AQ,BQ)=\measuredangle (AQ,PQ)+\measuredangle (PQ,BQ)=\measuredangle (AD,PD)+\measuredangle (PC,BC),\\ \measuredangle (CR,DR)& =\measuredangle (CQ,DQ)=\measuredangle (CQ,PQ)+\measuredangle (PQ,DQ)=\measuredangle (CB,PB)+\measuredangle (PA,DA),\\ \measuredangle (BR,CR)& =\measuredangle (BR,RQ)+\measuredangle (RQ,CR)=\measuredangle (BA,AQ)+\measuredangle (DQ,CD)\\ & =\measuredangle (BA,AP)+\measuredangle (AP,AQ)+\measuredangle (DQ,DP)+\measuredangle (DP,CD)\\ & =\measuredangle (BA,AP)+\measuredangle (DP,CD).\end{array}$$ Observe that the whole construction is reversible. One may start with point $R$, define $Q$ as the second intersection of circles ($ARB$) and ($CRD$), and then define $P$ as the second intersection of circles ($AQD$) and ($BQC$). The equalities above will still hold true. Assume in addition that $P$ is interior to $ABCD$. Then $$\begin{array}{rl}\measuredangle (AD,PD)=\angle PDA,\measuredangle (PC,BC)& =\angle PCB,\measuredangle (CB,PB)=\angle PBC,\measuredangle (PA,DA)=\angle PAD\\ \measuredangle (BA,AP)& =\angle PAB,\measuredangle (DP,CD)=\angle PDC.\end{array}$$

i. Suppose that $P$ lies in $ABCD$ and satisfies (1). Then $\measuredangle (AR,BR)=\angle PDA+\angle PCB=90^{\circ }$ and similarly $\measuredangle (BR,CR)=\measuredangle (CR,DR)=90^{\circ }$. It follows that $R$ is the common point of lines $AC$ and $BD$, and that these lines are perpendicular.

ii. Suppose that $AC$ and $BD$ are perpendicular and intersect at $R$. We show that the point $P$ defined by the reverse construction (starting with $R$ and ending with $P$) lies in $ABCD$. This is enough to finish the solution, because then the angle equalities above will imply (1). One can assume that $Q$, the second common point of circles ($ABR$) and ($CDR$), lies in $\angle ARD$. Then in fact $Q$ lies in triangle $ADR$ as angles $AQR$ and $DQR$ are obtuse. Hence $\angle AQD$ is obtuse, too, so that $B$ and $C$ are outside circle ($ADQ$) ($\angle ABD$ and $\angle ACD$ are acute). Now $\angle CAB+\angle CDB=\angle BQR+\angle CQR=\angle CQB$ implies $\angle CAB<\angle CQB$ and $\angle CDB<\angle CQB$. Hence $A$ and $D$ are outside circle ($BCQ$). In conclusion, the second common point $P$ of circles ($ADQ$) and ($BCQ$) lies on their $\mathrm{arcs}ADQ$ and $BCQ$. We can assume that $P$ lies in $\angle CQD$. Since $$\begin{array}{c}\angle QPC+\angle QPD=\left(180^{\circ }-\angle QBC\right)+\left(180^{\circ }-\angle QAD\right)=\\ =360^{\circ }-(\angle RBC+\angle QBR)-(\angle RAD-\angle QAR)=360^{\circ }-\angle RBC-\angle RAD>180^{\circ },\end{array}$$ point $P$ lies in triangle $CDQ$, and hence in $ABCD$. The proof is complete.